Choose $u_{m}\in W_{p}^{2}(\Omega)\cap W_{q}^{2}(\Omega)$ to harmonic function $u$.

Question

Let $\Omega$ : bounded, open set and $\frac{1}{p}+\frac{1}{q}=1$.

Let $u\in W_{p}^{2}(\Omega)$ satisfy $\triangle u=0$ on $\Omega$.

Then want to choose $u_{m}\in W_{p}^{2}(\Omega)\cap W_{q}^{2}(\Omega)$ such that $u_{m}\rightarrow u$ in $W_{p}^{2}(\Omega)$.

If $p<2$, then $W_{p}^{2}\cap W_{q}^{2}=W_{p}^{2}$. So just we can choose $u_{m}=u$.

If $p\geq2$, $W_{p}^{2}\cap W_{q}^{2} =W_{q}^{2}\subset W_{p}^{2}$. I can't go further here.

Answer 1

From the result in Evans' classical book, we know that if $\Omega$ is of $C^1$, then we can find $u_m\in C^\infty(\Omega)$ such that $u_m\to u$ in $W^{k,p}(\Omega)$.

Actually, there is a stronger density result, whose statement is as follows: